Examples of Creative Problem-Solving

[Gopal Mailpalli]

Can you list few examples.

 

[Dale]

Is this homework?

 

[Gopal Mailpalli]

Is this homework?

No dale, it isn't home work. While doing a problem (self-study), i came across this. So asked here.

 

[Dale]

Ok, you should still show some effort and thought about it on your own. How are velocity and acceleration defined?

 

[Gopal Mailpalli]

Ok, you should still show some effort and thought about it on your own. How are velocity and acceleration defined?

The rate of change of velocity is acceleration.

 

[Dale]

And what is velocity?

 

[Gopal Mailpalli]

And what is velocity?

Velocity is defined as the rate of change of position. Velocity is the first derivative of position with respect to time, where as acceleration is the second derivative of position.

 

[A.T.]

Velocity is the first derivative of position with respect to time, where as acceleration is the second derivative of position.

And how are those points called where the 1st derivative is zero, but the 2nd isn't.

 

[Dale]

Velocity is the first derivative of position with respect to time, where as acceleration is the second derivative of position.

Perfect. So if, for example, your position is $x=-t^2+3t+5$, then what is your velocity and acceleration?

 

[Gopal Mailpalli]

Perfect. So if, for example, your position is $x=-t^2+3t+5$, then what is your velocity and acceleration?

Velocity is -2t + 3 and acceleration is -2, with its respective units.

 

[Gopal Mailpalli]

And how are those points called where the 1st derivative is zero, but the 2nd isn't.

Pardon me, I didn't understand the question.

 

[Ibix]

Velocity is -2t + 3 and acceleration is -2, with its respective units.

So the answer to your original question is...?

 

[Dale]

Velocity is -2t + 3 and acceleration is -2, with its respective units.

Correct. So is there any t for which v=0? What is the acceleration at that time?

Also, plot the position as a function of time. Do you notice anything special about the time you found above?

 

[mathman]

Think of the motion of a pendulum.

 

[Gopal Mailpalli]

Think of the motion of a pendulum.

Thank you, i understood that at extreme positions, the velocity remains zero but acceleration is non-zero (changes its sign)

 

[Gopal Mailpalli]

Correct. So is there any t for which v=0? What is the acceleration at that time?

Also, plot the position as a function of time. Do you notice anything special about the time you found above?

For t = 3/2, the velocity is zero. Based on the graph, the position of the object is constant w.r.t time. How would one determine the acceleration then?

 

[Dale]

Well done!

You already found the acceleration above, a=-2, regardless of time. Visually, a negative acceleration gives a position graph which is concave down, and a positive acceleration gives a position graph which is concave up. Since this graph is concave down everywhere you can immediately tell that the acceleration is negative everywhere.

 

[A.T.]

Thank you, i understood that at extreme positions, the velocity remains zero

Velocity doesn't remain zero, because acceleration is not zero. Velocity is instantaneously zero.

but acceleration is non-zero (changes its sign)

The velocity changes its sign, not the acceleration.